A miniature roulette wheel is divided into 10 equal sectors

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

A miniature roulette wheel is divided into 10 equal sectors

by BTGmoderatorLU » Wed Jan 24, 2018 5:24 am

A miniature roulette is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly detemines the winning sector by setting in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors' integers will be even?

A. 88%
B. 75%
C. 67%
D. 63%
E. 50%

The OA is A.

I'm really confused with this PS question. Experts, any suggestion about how can I solve this question? Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Wed Jan 24, 2018 5:24 am

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Wed Jan 24, 2018 7:28 pm

Hi LUANDATO,

We're told that a roulette wheel is divided into 10 equal sectors, with each sector bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly detemines the winning sector by setting in that sector. The wheel is spun three times and we're asked for the approximate probability that the PRODUCT of the three winning sectors' integers will be EVEN. This question is based on some Number Properties and Probability math.

To start, the product of any integer and an EVEN number is EVEN. Thus, if any of the 3 numbers is EVEN, then the product will be EVEN. Rather than calculate all of the different outcomes that involve at least one even number, we can instead calculate what we DON'T WANT to have happen (the product is ODD); we can then subtract that probability from the number 1 to determine the probability of what we DO want (the product is EVEN).

With the integers 1-10 inclusive, we have 10 total values (5 odds and 5 evens). The probability of the three numbers all being ODD is....

(Odd)(Odd)(Odd) = (5/10)(5/10)(5/10) = (1/2)(1/2)(1/2) = 1/8

Thus, the probability of AT LEAST one of the three numbers being EVEN is 1 - 1/8 = 7/8

1/8 = .125 = 12.5%
7/8 = .875 = 87.5%

The approximate value of our calculation is 88%

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Fri Jan 26, 2018 12:26 pm

LUANDATO wrote:A miniature roulette is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly detemines the winning sector by setting in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors' integers will be even?

A. 88%
B. 75%
C. 67%
D. 63%
E. 50%

We can use the formula:

P(even product) = 1 - P(odd product)

To get an odd product we must have all odd numbers, to the probability of 3 odds is:

1/2 x 1/2 x 1/2 = 1/8.

So P(even product) = 1 - 1/8 = 7/8 â‰ˆ 88%

Answer: A

Jeffrey Miller
Head of GMAT Instruction
[email protected]